
A p-n photodiode is fabricated from a semiconductor with a band gap of $\text{2}\text{.8eV}$. Can it detect a wavelength of $\text{6000nm}$?
Answer
556.5k+ views
Hint: The band gap of a semiconductor represents the minimum amount of energy required to jump an electron from valence band to the conduction band. The energy of a photon is given as the product of the Planck’s constant and the frequency of the incident light and the frequency is equal to the ratio of speed of light and the wavelength of light. Using this formula, we can obtain the required answer.
Complete answer:
The band gap between the valence band and the conduction band of a semiconductor is known as the band gap of the semiconductor. The band gap is greater in case of insulators, a little closer than the case of insulators for semiconductors, and the band gap is almost or completely absent in the case of a conductor.
Now, to solve the above question, we will require to use the very basic formula for the and gap energy in a semiconductor.
So, now, let us consider that ${{\text{E}}_{\text{g}}}$is the band gap of the semiconductor.
Therefore, we can write the formula for ${{\text{E}}_{\text{g}}}$as,
${{E}_{g}}=h\upsilon $
where, $\text{h}$is the Planck’s constant, $\text{ }\!\!\upsilon\!\!\text{ }$is the threshold frequency and $\text{c}$ is the velocity of light.
Now,
$\begin{align}
& \upsilon =\dfrac{c}{\lambda } \\
& \Rightarrow h\upsilon =\dfrac{hc}{\lambda } \\
\end{align}$
Therefore,
${{E}_{g}}=\dfrac{hc}{\lambda }$
Now, let us substitute the given values and determine the maximum wavelength ,
$\begin{align}
& 2.8\times 1.602\times {{10}^{-19}}=\dfrac{6.626\times {{10}^{-34}}\times 3\times {{10}^{8}}}{\lambda } \\
& \Rightarrow \lambda =443.15nm \\
\end{align}$
So, the maximum wavelength that can be detected by the semiconductor is $\text{443}\text{.15nm}$.
Therefore, we can conclude that the semiconductor cannot detect a wavelength of $\text{6000nm}$.
Note:
The calculated wavelength is taken to be maximum because the formula used is for threshold frequency which is the smallest frequency, and because frequency and wavelength are inversely proportional to each other, so the corresponding wavelength is maximum.
Complete answer:
The band gap between the valence band and the conduction band of a semiconductor is known as the band gap of the semiconductor. The band gap is greater in case of insulators, a little closer than the case of insulators for semiconductors, and the band gap is almost or completely absent in the case of a conductor.
Now, to solve the above question, we will require to use the very basic formula for the and gap energy in a semiconductor.
So, now, let us consider that ${{\text{E}}_{\text{g}}}$is the band gap of the semiconductor.
Therefore, we can write the formula for ${{\text{E}}_{\text{g}}}$as,
${{E}_{g}}=h\upsilon $
where, $\text{h}$is the Planck’s constant, $\text{ }\!\!\upsilon\!\!\text{ }$is the threshold frequency and $\text{c}$ is the velocity of light.
Now,
$\begin{align}
& \upsilon =\dfrac{c}{\lambda } \\
& \Rightarrow h\upsilon =\dfrac{hc}{\lambda } \\
\end{align}$
Therefore,
${{E}_{g}}=\dfrac{hc}{\lambda }$
Now, let us substitute the given values and determine the maximum wavelength ,
$\begin{align}
& 2.8\times 1.602\times {{10}^{-19}}=\dfrac{6.626\times {{10}^{-34}}\times 3\times {{10}^{8}}}{\lambda } \\
& \Rightarrow \lambda =443.15nm \\
\end{align}$
So, the maximum wavelength that can be detected by the semiconductor is $\text{443}\text{.15nm}$.
Therefore, we can conclude that the semiconductor cannot detect a wavelength of $\text{6000nm}$.
Note:
The calculated wavelength is taken to be maximum because the formula used is for threshold frequency which is the smallest frequency, and because frequency and wavelength are inversely proportional to each other, so the corresponding wavelength is maximum.
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